FloatingPointRepresentation

Clearly x b0b1b2 b232 2e thus we have for any

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . Clearly, x; = (b0:b1b2 : : :b23)2 2E : Thus we have, for any rounding mode, that jround(x) ; xj < 2;23 2E while for round to nearest jround(x) ; xj 2;24 2E : Similar results hold for double and extended precision, replacing 2;23 by 2;52 and 2;63 respectively, so that in general we have jround(x) ; xj < 2E (1) for any rounding mode and 1 jround(x) ; xj 2 2E for round to nearest. round towards zero, could the absolute rounding error be exactly equal to 2E ? For round to nearest, could the absolute rounding 1 error be exactly equal to 2 2E ? Exercise 10 For Exercise 11 Does (1) hold if x is subnormal, i.e. E = ;126 and b0 = 0? The presence of the factor 2E is inconvenient, so let us consider the relative rounding error associated with x, de ned to be = round(x) ; 1 = round(x) ; x : x x 14 Since for normalized numbers x = m 2E where m 1 (because b0 = 1) we have, for all rounding modes, E j j < 2E2 = : (2) In the case of round to nearest, we have jj Exercise 12 Does (2) hold if If not, how big could be? 1 2 2...
View Full Document

Ask a homework question - tutors are online